If the mid-points of the sides \(B C, C A\) and \(A B\) of a triangle \(A B C\), are respectively \((2,1)\),…

If the mid-points of the sides \(B C, C A\) and \(A B\) of a triangle \(A B C\), are respectively \((2,1)\), \((-1,-2)\) and \((3,3)\), then the equation of the side \(B C\) is
  1. \(x-2 y=0\)
  2. \(5 x-4 y=6\)
  3. \(2 x+3 y=8\)
  4. \(3 x-2 y=6\)

Solution


Let the coordinate of \(B\) is \((a, b)\). Mid-point of \(B Q=\) Mid-point of \(P R\) \(\begin{aligned} & \Rightarrow \quad\left(\frac{a-1}{2}, \frac{b-2}{2}\right)=\left(\frac{3+2}{2}, \frac{3+1}{2}\right) \\ & \Rightarrow \quad \frac{a-1}{2}=\frac{5}{2} \\ & \text{and } \frac{b-2}{2}=\frac{4}{2} \\ & \Rightarrow \quad a=5+1 \\ & \text{and } b=4+2 \\ & \Rightarrow \quad a=6 \text { and } b=6 \end{aligned}\) \(\therefore \quad B(6,6)\) and \(P(2,1)\) Now, equation of \(B C\) is \(\begin{array}{rlrl} & y-6=\frac{1-6}{2-6}(x-6) \\ \Rightarrow & y-6=\frac{-5}{-4}(x-6) \\ \Rightarrow & 4 y-24=5 x-30 \\ \Rightarrow & 5 x-4 y-30+24=0 \\ \Rightarrow & 5 x-4 y=6 \end{array}\)

Asked in: AP EAMCET 2019 (22 Apr Shift 1)

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